Difficulty: Advanced | Prerequisites: All derivative rules, critical points, and MVT (Ch. 2 and 3.1–3.2 notes).
This set covers Stewart sections 3.3 through 3.8. These sections put derivatives to work in the most exam-heavy ways: determining where a graph is rising or falling, concave up or down, and locating inflection points; evaluating tricky limits with L'Hopital's rule; drawing complete curve sketches; solving real-world optimisation problems; and using Newton's method to find roots numerically.
The first and second derivatives reveal the complete shape of a graph: where it rises, falls, bends upward, bends downward, and has inflection points. L'Hopital's rule handles indeterminate limits (0/0 and infinity/infinity) by replacing the original limit with the limit of the ratio of derivatives. Optimisation problems translate real-world constraints into a function to maximise or minimise. Newton's method uses tangent lines to zero in on roots numerically.
First derivative test
If f' changes from positive to negative at a critical point c, then f has a local max at c. If f' changes from negative to positive, f has a local min. If f' does not change sign, c is neither.
Second derivative test
If f'(c) = 0 and f''(c) > 0, then f has a local min at c. If f''(c) < 0, local max. If f''(c) = 0, the test is inconclusive.
Concavity
A function is concave up where f''(x) > 0 (the graph bends upward, like a cup). It is concave down where f''(x) < 0 (the graph bends downward, like a cap).
Inflection point
A point where the concavity changes, from up to down or vice versa. At an inflection point, f''(x) = 0 or f''(x) does not exist, but not every such point is an inflection point.
L'Hopital's rule
If lim f(x)/g(x) gives 0/0 or infinity/infinity, then lim f(x)/g(x) = lim f'(x)/g'(x), provided the latter limit exists.
Newton's method
An iterative root-finding algorithm: starting from a guess x_0, compute x_{n+1} = x_n - f(x_n)/f'(x_n). Each step follows the tangent line to a new, better approximation of the root.
Increasing/decreasing test: f is increasing where f'(x) > 0 and decreasing where f'(x) < 0. Find the critical points, then test the sign of f' in each interval between them
First derivative test: at a critical point c, if f' switches from + to -, f has a local max. If f' switches from - to +, f has a local min. No sign change means no extremum
Concavity test: f is concave up where f''(x) > 0, concave down where f''(x) < 0
Second derivative test: if f'(c) = 0 and f''(c) > 0, then c is a local min (concave up, so the horizontal tangent sits at the bottom of a bowl). If f''(c) < 0, c is a local max. If f''(c) = 0, the test is inconclusive and you fall back to the first derivative test
Inflection points occur where the concavity changes. Find candidates by solving f''(x) = 0 or where f'' is undefined, then verify that f'' actually changes sign
Applies when lim f(x)/g(x) is an indeterminate form 0/0 or infinity/infinity
The rule: lim f(x)/g(x) = lim f'(x)/g'(x), provided the right-hand limit exists (or is infinity)
You may apply L'Hopital's rule repeatedly if the result is still indeterminate
Example: lim(x->0) sin(x)/x = lim(x->0) cos(x)/1 = 1
Example: lim(x->infinity) x^2/e^x = lim 2x/e^x = lim 2/e^x = 0 (applied twice)
Other indeterminate forms (0 * infinity, infinity - infinity, 0^0, 1^infinity, infinity^0) must first be rewritten as a 0/0 or infinity/infinity quotient before applying the rule
Warning: L'Hopital's rule does not apply when the form is not indeterminate. Applying it to a non-indeterminate form gives a wrong answer
A complete curve sketch combines all derivative information into one picture. The standard checklist:
Domain: where is f(x) defined?
Intercepts: set y = 0 for x-intercepts, x = 0 for y-intercept
Symmetry: is f even (f(-x) = f(x)), odd (f(-x) = -f(x)), or periodic?
Asymptotes: vertical (denominator = 0), horizontal (lim as x -> +/- infinity), oblique (if degree of numerator exceeds denominator by 1)
First derivative analysis: intervals of increase/decrease, local maxima and minima
Second derivative analysis: intervals of concavity, inflection points
Plot key points and sketch
Horizontal asymptotes: lim(x->infinity) f(x) and lim(x->-infinity) f(x). For rational functions, compare leading-term degrees
Procedure:
Read the problem and identify the quantity to be maximised or minimised
Draw a picture and assign variables
Write the objective function (the thing to optimise) in terms of those variables
Use the constraint(s) to reduce to a single-variable function
Find critical points by setting the derivative to zero
Confirm your answer is a max or min (using the closed interval method, or first/second derivative test)
Common setups: maximising area given a fixed perimeter, minimising material for a container of fixed volume, maximising profit given cost and revenue functions
Always state the domain of your single-variable function. Physical constraints (lengths must be positive, etc.) restrict the domain and can provide endpoints to check
Goal: find a root of f(x) = 0 numerically
Iteration formula: x_{n+1} = x_n - f(x_n) / f'(x_n)
Start with an initial guess x_0, then iterate. Each step follows the tangent line at (x_n, f(x_n)) down to the x-axis
Example: find sqrt(2) by solving x^2 - 2 = 0. With x_0 = 1: x_1 = 1 - (1-2)/(2*1) = 1.5, x_2 = 1.5 - (2.25-2)/(3) = 1.4167, x_3 = 1.41422..., converging rapidly
Convergence is typically very fast (quadratic) when the starting guess is close to the root and f'(x) is nonzero at the root
Failure modes: if f'(x_n) = 0, the tangent is horizontal and the method breaks. A bad initial guess can send the iterations away from the intended root or into a cycle
Students often assume f''(c) = 0 means c is an inflection point. It does not, unless f'' actually changes sign there. For example, f(x) = x^4 has f''(0) = 0, but x = 0 is a local min, not an inflection point.
Students apply L'Hopital's rule to forms that are not indeterminate (e.g. 1/0 or 5/3). The rule only applies to 0/0 or infinity/infinity. Using it otherwise gives a wrong answer.
Students forget to check endpoints and domain restrictions in optimisation problems. A critical point in the interior may not be the global optimum if an endpoint gives a larger (or smaller) value.
Students confuse Newton's method with the bisection method. Newton's method uses the tangent line and converges much faster, but requires f'(x) and can fail with a bad starting guess. Bisection is slower but more robust.
⚠️ The first and second derivative tests for local extrema are tested on nearly every exam. Know both, and know when the second derivative test is inconclusive (f''(c) = 0).
⚠️ Full curve sketching problems (domain, intercepts, asymptotes, increase/decrease, concavity, key points) are classic exam questions worth many marks.
⚠️ L'Hopital's rule: always verify the form is indeterminate before applying. Show this verification in your working.
⚠️ Optimisation word problems: the most common Calc I exam type. Practise fencing problems, box problems, and distance-minimisation problems until the setup step is automatic.
⚠️ Newton's method: you may be asked to perform two or three iterations from a given starting point.
If f'(c) = 0 and f''(c) > 0, is c a local max or local min? (Local min.)
True or false: f''(x) = 0 always means an inflection point. (False. f'' must change sign.)
What indeterminate forms does L'Hopital's rule handle directly? (0/0 and infinity/infinity.)
In Newton's method, x_{n+1} = x_n - ____. (f(x_n) / f'(x_n).)
True or false: in an optimisation problem on a closed interval, the absolute max can occur at an endpoint. (True.)
Q: Find the intervals of increase, decrease, and the local extrema of f(x) = x^3 - 12x + 1.
A: f'(x) = 3x^2 - 12 = 3(x-2)(x+2). Critical points: x = -2, x = 2. Sign chart: f' > 0 on (-infinity, -2), f' < 0 on (-2, 2), f' > 0 on (2, infinity). So f is increasing on (-infinity, -2) and (2, infinity), decreasing on (-2, 2). Local max at x = -2 (f(-2) = 17), local min at x = 2 (f(2) = -15).
Q: Evaluate lim(x->0) (e^x - 1 - x) / x^2.
A: At x = 0 this is 0/0. L'Hopital: lim (e^x - 1) / (2x), still 0/0. L'Hopital again: lim e^x / 2 = 1/2.
Q: A farmer has 200 m of fencing and wants to enclose a rectangular area against a river (no fence needed on the river side). What dimensions maximise the area?
A: Let x = width (two sides), y = length (one side). Constraint: 2x + y = 200, so y = 200 - 2x. Objective: A = xy = x(200 - 2x) = 200x - 2x^2. A'(x) = 200 - 4x = 0 gives x = 50. Then y = 100. A''(x) = -4 < 0, confirming a max. Maximum area = 5000 m^2.
Q: Use Newton's method with x_0 = 2 to approximate a root of f(x) = x^3 - 5.
A: f'(x) = 3x^2. x_1 = 2 - (8-5)/(12) = 2 - 0.25 = 1.75. x_2 = 1.75 - (5.359-5)/(9.1875) = 1.75 - 0.039 = 1.7109. The cube root of 5 is approximately 1.7100.
The increasing/decreasing test and concavity analysis from this chapter use the MVT results from Ch. 3.2. L'Hopital's rule reappears frequently when evaluating improper integrals in Calculus II. The optimisation framework extends to multivariable calculus (Lagrange multipliers) in Calculus III. Newton's method is a cornerstone of numerical analysis and is used in scientific computing, engineering simulations, and machine learning optimisation.
first derivative test, second derivative test, concavity, concave up, concave down, inflection point, L'Hopital's rule, indeterminate form, 0/0, infinity/infinity, curve sketching, asymptotes, horizontal asymptote, optimisation, optimization, maximise, minimise, fencing problem, Newton's method, Newton-Raphson, root finding, iteration, MATH 132, Stewart Chapter 3, Calculus I